Video: Integrating a Trigonometric Function Whose Argument Has the Form π‘Žπ‘₯ + 𝑏

Determine ∫cos (7π‘₯ βˆ’ 7) dπ‘₯.

03:31

Video Transcript

Determine the integral of the cos of seven π‘₯ minus seven with respect to π‘₯.

The question is asking us to evaluate the integral of a composite function. It’s the integral of the composition of a trigonometric function and a linear function. And there’s a lot of different ways we could think about evaluating this integral. For example, we might want to try using our multiple angle formulas to rewrite our integrand. However, this is also the integral of a composite function. So we should also check if there’s a suitable 𝑒 substitution we can use to make this integral easier. And in this case, this is true. We’ll set 𝑒 to be our inner function. That’s the linear function seven π‘₯ minus seven. Then, we can differentiate 𝑒 with respect to π‘₯. Since this is a linear function, we just get the coefficient of π‘₯ which is seven.

At this point, we need to be a little bit careful. Normally, we would check to see if d𝑒 by dπ‘₯ appears in our integrand so we could use this to make or integral simpler. However, on first glance, this doesn’t seem to be the case. However, if we multiply our integrand by seven and then multiply our entire integral by one over seven, we don’t change the value of our integral. But now we have d𝑒 by dπ‘₯ appearing in our integrand. And in doing this, we’ve rewritten our integral to be of the form one-seventh times the integral of 𝑔 prime of π‘₯ times 𝑓 of 𝑔 of π‘₯ with respect to π‘₯. We can use the reverse chain rule.

So let’s try and evaluate our integral by using the substitution 𝑒 is equal to seven π‘₯ minus seven. We’ve already shown d𝑒 by dπ‘₯ is equal to seven. And remember, d𝑒 by dπ‘₯ is not a fraction. However, when we’re integrating by substitution, we can treat it a little bit like a fraction. This gives us the equivalent statement in terms of differentials d𝑒 is equal to seven dπ‘₯. We’re now ready to evaluate our integral by using substitution. First, the integral of the cos of seven π‘₯ minus seven with respect to π‘₯ is equal to one-seventh times the integral of seven times the cos of seven π‘₯ minus seven with respect to π‘₯.

Next, we set 𝑒 to be our inner function seven π‘₯ minus seven. We then show that d𝑒 by dπ‘₯ is equal to seven. And this is equivalent to saying d𝑒 is equal to seven dπ‘₯. In other words, in our integration by substitution, we can replace seven dπ‘₯ with d𝑒. So using integration by substitution, we get one-seventh multiplied by the integral of the cos of 𝑒 with respect to 𝑒. And we know how to evaluate this integral. The integral of the cos of πœƒ with respect to πœƒ is equal to the sin of πœƒ plus the constant of integration 𝑐. So applying this rule with 𝑒 instead of πœƒ, we get one-seventh times the sin of 𝑒 plus the constant of integration we’ll call 𝑐 one.

Distributing one-seventh over our parentheses, we get one-seventh times the sin of 𝑒 plus 𝑐 one over seven. And we’ll simplify our answer. First, 𝑐 one is a constant. So 𝑐 one divided by seven is also a constant. So we’ll just replace this with a new constant we’ll call 𝑐. Next, remember, our original integral was in terms of π‘₯. So we want to give our answer in terms of π‘₯. We’ll do this by reusing our substitution 𝑒 is equal to seven π‘₯ minus seven. This gives us one-seventh times the sin of seven π‘₯ minus seven plus 𝑐. And this is our final answer.

Therefore, by using integration by substitution, we’ve shown the integral of the cos of seven π‘₯ minus seven with respect to π‘₯ is equal to one-seventh times the sin of seven π‘₯ minus seven plus our constant of integration 𝑐.

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